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662 CHAPTER 18 The Potential Equation

Now we need to choose the c n ’s so that
∞
∂u nπ nπa nπ
(a, y) = g(y) = c n sinh cos y .
∂x b b b
n=1
This is a Fourier cosine expansion of g(y) on [0,b]. Notice that the constant term in this
expansion is zero. But this constant term is
1 b
g(y)dy,
b 0
and we would have a contradiction if this integral were not zero. In this event, this problem would
have no solution.
For the other coefficients in this cosine expansion, we have
nπ nπa 2 b nπξ
c n sinh = g(ξ)cos dξ,
b b b 0 b
so
2 b nπξ
c n = g(ξ)cos dξ.
nπ sinh(nπa/b) 0 b
With this choice of the coefficients, the solution of the Neumann problem is
∞

nπx nπy
u(x, y) = c 0 + c n cosh cos .
b b
n=1
Here c 0 is an arbitrary constant. Neumann problems do not have unique solutions: If u is a
solution, so is u + c for any number c.
18.8.2 A Neumann Problem for a Disk
Suppose D is a disk of radius R about the origin. The boundary is the circle C of radius R about
the origin. In polar coordinates, the Neumann problem for D is
2
∇ u(r,θ) = 0for 0 ≤r < R,−π ≤ θ ≤ π
and
∂u
(R,θ) = f (θ) for − π ≤ θ ≤ π.
∂r
Notice that the normal derivative to C is ∂u/∂r because the line from the origin to a point of C
is perpendicular to C at that point.
A necessary condition for existence of a solution is that
π
f (θ)dθ = 0,
−π
which is a condition we will assume for f (θ).
Attempt a solution
∞
1
n
n
u(r,θ) = a 0 + [a n r cos(nθ) + b n r sin(nθ)].
2
n=1
We must choose the coefficients to satisfy
∂u
(R,θ) = f (θ)
∂r
∞
n−1 n−1
= [na n R cos(nθ) + nb n R sin(nθ)].
n=1

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October 14, 2010 15:27 THM/NEIL Page-662 27410_18_ch18_p641-666